Localization Let A be a commutative ring. By a multiplicative set S ⊂ A, we mean a subset such that 1 2 S −1 −1 and if s; t 2 S then st 2 S. We will construct a ring S A and a homomorphism iS : A ! S A that has the following universal property: −1 For all s 2 S, iS(s) is invertible in S A, and for every ring homomorphism φ : A ! B such −1 that φ(s) is invertible in B for all s 2 S, there exists a unique homomorphism φS : S A ! B such −1 that φ = φS ◦ iS : A ! S A ! B. In the definition of multiplicative set, 0 2 S is technically allowed. However, it is very obvious in that case that S−1A = (0), the zero ring, since 0 invertible implies 1 = 0. This is not very interesting. If 1 6= 0 in A, that is, if A 6= (0), and if 0 62 S, then S−1A 6= (0) will follow as well. If A is an integral domain and 0 62 S, then S−1A is a rather obvious subring of the field of fractions of A, namely, the subring of fractions whose denominators are elements of S. But even in this case the main point is not just the construction of S−1A, but rather the connections between ideals in A and S−1A and the many applications in commutative algebra, number theory, and algebraic geometry. The usual argument with universal properties shows that if S−1A exists, it is unique up to isomorphism. The ring S−1A is called a localization of A. This terminology arises from consideration of rings of continuous functions, with values, say, in a field. If A is a ring of functions on a space X and if Y ⊂ X is a subspace, which could be a single point, let S = S(Y ) ⊂ A denote the subset of functions which have no zeros on Y . It is clear that S is a multiplicative set. The quotients a=s, with a 2 A and s 2 S, represent well-defined germs of functions near Y . That is, each such quotient defines a function in some neighborhood of Y , specifically, the neighborhood where s is non-zero. These germs can be added and multiplied and form a ring, whose algebraic properties reflect the properties of X locally, that is, near Y . The ring A might have zero divisors, some elements of S might even be zero divisors, but the ring of germs of functions near Y which can be expressed as quotients a=s, still makes sense. It is perhaps useful to keep this picture in mind as motivation for the general construction below. Before constructing S−1A, we list some examples of multiplicative sets. (i) S is the set of all non-zero divisors in A. If A is an integral domain, this is just the set of all non-zero elements of A, and S−1A will turn out to be the field of fractions of A. (ii) S = A − P , where P ⊂ A is a prime ideal. In this situation, the localization S−1A is often denoted A(P ), and referred to as the localization of A at the prime P . The ring A(P ) is a local ring, that is, a ring with a unique maximal ideal. If A = A(V ) is the affine coordinate ring of a variety over an algebraically closed field K, and if P = I(p) ⊂ A is the maximal ideal corresponding to a point p 2 V , then the localization A(P ) is a ring of germs of K-valued functions defined on open neighborhoods of p 2 V in the Zariski topology. (It is a little harder to interpret the localization A(P ) as a ring of functions if P = I(W ) ⊂ A is a prime ideal corresponding to an irreducible subvariety W ⊂ V .) (iii) It is obvious that arbitrary intersections of multiplicative sets are multiplicative sets. For example, if fPjg is a family of prime ideals in A, then \j(A−Pj) = A−[jPj is a multiplicative set. (iv) If s 2 A is an element which is not nilpotent, then S = fsn j n ≥ 0g is a multiplicative set. Suppose φ : A ! B is a ring homomorphism so that some element s 2 S is a zero divisor in A and φ(s) 2 B is invertible. If as = 0 2 A then obviously φ(a) = 0 2 B, since φ(a)φ(s) = 0. This −1 observation makes it clear that iS : A ! S A will not be injective if any elements of S are zero- divisors in A. A consequence is that the construction of S−1A is somewhat more subtle than, say, 1 the construction of the field of fractions of an integral domain. On the other hand, this observation about consequences of the existence of zero divisors in S motivates the precise definition of S−1A in general, which we now give. S−1A = f[a=s] j a 2 A; s 2 Sg=∼, where [a0=s0] ∼ [a00=s00] if and only if there exists s 2 S with a0s00s = a00s0s 2 A. −1 The homomorphism iS : A ! S A will be defined by iS(a) = [a=1]. The symbol [a=s] will −1 −1 0 0 0 00 0 00 0 00 0 00 name the element iS(a)iS(s) 2 S A. So we certainly want [a =s ] = [a s =s s ] = [a s s=s s s], and [a00=s00] = [a00s0=s00s0] = [a00s0s=s00s0s]. The usual definition of equality of fractions, a0=s0 = a00=s00 if a0s00 = a00s0, would be adequate if no element of S were a zero divisor. The additional factor of s in the definition above allows for the possibility that a0s00 − a00s0 6= 0, but a0s00s − a00s0s = 0 for some s 2 S. Exercise 1 Define addition and multiplication in S−1A, and verify that these operations are well- defined. Acknowledge that you ought to also check all ring axioms, associative laws, distributive laws, identity elements, etc, but don't bother with all that. −1 Exercise 2 Verify the universal property, stated at the outset, for iS : A ! S A. −1 Exercise 3 Characterize the set of all elements t 2 A such that iS(t) is invertible in S A. (This subset of A is also a multiplicative set, called the saturation of S in A. Localizing A with respect to the saturation of S yields the same ring as localizing with respect to S. This can be seen instantly, since the universal property of S−1A implies that this ring also has the required universal property to be the localization with respect to the saturation of S.) Next, we investigate relations between localization and ideals. Suppose I ⊂ A is an ideal with I \ S = ;. Let A∗ = A=I and let S∗ be the image of S in A∗. Then S∗ is a multiplicative set in A∗. The following exercise expresses the commutativity of localization and residue ring formation. One can either divide by an ideal first and then localize, or localize first and then divide by an appropriate ideal. Exercise 4 (S∗)−1A∗ =∼ S−1A=IS−1A. (Although it is easy enough to establish one-to-one ring homomorphisms in both directions, it is more elegant to verify that the ring on the right has the universal property of the ring on the left, hence it is isomorphic to the ring on the left.) If φ : A ! B is any ring homomorphism there are operations known as contraction and extension which relate ideals in B and A. If J ⊂ B is an ideal, the contraction is J c = φ−1(J) ⊂ A, the inverse image of J in A. If I ⊂ A is an ideal, the extension is Ie = φ(I)B, the ideal generated by the image of I in B. Both the contraction and extension operations preserve inclusions, that is, J1 ⊂ J2 c c ce ec implies J1 ⊂ J2, and similarly for extension. There are also obvious inclusions J ⊂ J and I ⊂ I , from which follow trivially the relations J cec = J c and Iece = Ie. The contraction operation has a number of reasonable properties, but very little can be said in general about extension. For example, the following properties of contraction are all easy to check, while the analogues for extension are all false. (i) If J is prime, primary, or radical, then J c is, respectively, prime, primary, or radical. c c c c (ii) rad J = (rad J) and \(Ji ) = (\Ji) . −1 In case the homomorphism is a localization iS : A ! S A, one can say more about the contraction operation, and, in addition, the extension operation has many good properties. −1 ce e Exercise 5 If JS ⊂ S A is an ideal, then JS = JS. If I ⊂ A is an ideal, then I = f[a=s] j a 2 I; s 2 Sg ⊂ S−1A and Iec = fb 2 A j sb 2 I; for some s 2 Sg. 2 Exercise 6 Ie ⊂ S−1A is proper if and only if I \ S = ;. If I ⊂ A is prime or primary and I \ S = ;, then Ie ⊂ S−1A is, respectively, prime or primary. Moreover, in each of these cases, Iec = I ⊂ A. e e −1 Exercise 7 If I ⊂ A is any ideal, then rad I = (rad I) ⊂ S A. If I1 and I2 are ideals in A, e e e −1 then I1 \ I2 = (I1 \ I2) ⊂ S A.